Question

представьте многочлены в виде куба двучлена;

1)x^3+3x^2+3x+1

2)y^3-3y^2+3y-1

3)8-12p+6p^2-p^3

4)1+6q+12q^2+8q^3

5)125+75a+15a^2+a^3

6)0,008-0,12p+0,6p^2-p^3

Answer 1

$x^3+3x^2+3x+1=(x+1)^3$<var />

<var>$y^3-3y^2+3y-1=(y-1)^3$<var /></var>

<var>$8-12p+6p^2-p^3=(2-p)^3$<var /></var>

<var>$1+6q+12q^2+8q^3=(1+2q)^3$<var /></var>

<var>$125+75a+15a^2+a^3=(5+a)^3$<var /></var>

<var>$0,008-0,12p+0,6p^2-p^3=(0.2-p)^3$</var>

Answer 2

$1)x^3+3x^2+3x+1=(x+1)(x^2-x+1)+3x(x+1)=(x+1)(x^2-x+1+3x)=(x+1)(x^2+2x+1)=(x+1)^3\\\\ 2)y^3-3y^2+3y-1=(y-1)(y^2+y+1)-3y(y-1)=(y-1)(y^2+y+1-3y)=(y-1)(y^2-2y+1)=(y-1)^3\\\\ 3)8-12p+6p^2-p^3=-(p-2)^3\\\\ 4)1+6q+12q^2+8q^3=(2q+1)(4q^2-2q+1)+6q(1+2q)=(1+2q)(4q^2+4q+1)=(1+2q)^3\\\\ 5)125+75a+15a^2+a^3=(a+5)(a^2-5a+25)+15a(a+5)=(a+5)(a^2+10a+25)=(5+a)^3\\\\ 6)0,008-0,12p+0,6p^2-p^3=(0,2-p)^3$